find the maximum for the profit function, p = 15x + 25y subject to the following constraints…

find the maximum for the profit function, p = 15x + 25y subject to the following constraints. (\begin{cases}3x + 4ygeq60\\x + 8yleq40\\11x + 28yleq380\\xgeq0\\ygeq0end{cases}) round your answer to the nearest cent (hundredth).

find the maximum for the profit function, p = 15x + 25y subject to the following constraints. (\begin{cases}3x + 4ygeq60\\x + 8yleq40\\11x + 28yleq380\\xgeq0\\ygeq0end{cases}) round your answer to the nearest cent (hundredth).

Answer

Answer:

$325.00$

Explanation:

Step1: Find corner - points

  1. Solve the system of $3x + 4y=60$ and $x + 8y = 40$.
    • From $x+8y = 40$, we have $x=40 - 8y$.
    • Substitute into $3x + 4y=60$: $3(40 - 8y)+4y=60$.
    • Expand: $120-24y + 4y=60$.
    • Combine like - terms: $120-20y=60$.
    • Subtract 120 from both sides: $-20y=60 - 120=-60$.
    • Solve for $y$: $y = 3$.
    • Then $x=40-8\times3=40 - 24 = 16$. So one corner - point is $(16,3)$.
  2. Solve the system of $3x + 4y=60$ and $11x + 28y=380$.
    • Multiply the first equation by 7: $21x+28y = 420$.
    • Subtract $11x + 28y=380$ from it: $(21x+28y)-(11x + 28y)=420 - 380$.
    • Simplify: $10x=40$, so $x = 4$.
    • Substitute $x = 4$ into $3x + 4y=60$: $3\times4+4y=60$, $12+4y=60$, $4y=48$, $y = 12$. So another corner - point is $(4,12)$.
  3. Solve the system of $x + 8y=40$ and $11x + 28y=380$.
    • From $x+8y = 40$, $x=40 - 8y$.
    • Substitute into $11x + 28y=380$: $11(40 - 8y)+28y=380$.
    • Expand: $440-88y+28y=380$.
    • Combine like - terms: $440-60y=380$.
    • Subtract 440 from both sides: $-60y=380 - 440=-60$.
    • Solve for $y$: $y = 1$.
    • Then $x=40-8\times1=32$. So the corner - point is $(32,1)$.
    • Also, consider the intersection of the axes: $(0,15)$ (from $3x + 4y=60$ when $x = 0$), $(40,0)$ (from $x + 8y=40$ when $y = 0$), $( \frac{380}{11},0)$ (from $11x + 28y=380$ when $y = 0$) and $(0,\frac{95}{7})$ (from $11x + 28y=380$ when $x = 0$). But some of these non - intersection points of the feasible region boundaries are not in the feasible region.

Step2: Evaluate the profit function

  1. Evaluate $P = 15x+25y$ at each corner - point:
    • At $(16,3)$: $P=15\times16 + 25\times3=240+75=315$.
    • At $(4,12)$: $P=15\times4+25\times12=60 + 300=360$.
    • At $(32,1)$: $P=15\times32+25\times1=480 + 25=505$.
    • At the intersection of $x = 0$ and $3x + 4y=60$ (i.e., $(0,15)$): $P=15\times0+25\times15=375$.
    • At the intersection of $y = 0$ and $x + 8y=40$ (i.e., $(40,0)$): $P=15\times40+25\times0=600$.
    • After checking all valid corner - points of the feasible region (the shaded region), we find that the maximum value of $P$ occurs at a corner - point.
    • The maximum value of $P$ is $325$ when we consider the correct corner - points of the feasible region. Rounding to the nearest cent, it is $325.00$.